Pricing a call option with payoff function $C=\max\{S_T - S_{T/2}, 0\}$, where $S_T$ is geometric brownian motion. I appreciate any help! Please close this question if this is a duplicated question. Thanks all!

My approach is to take out $S_{T/2}$:

$$C = S_{T/2} \max\left\{\frac{S_T}{S_{T/2}} - 1, 0\right\}$$

Then we can define a risk neutral measure:

$$\begin{align} E[C] & = E\left[S_{T/2} \max\left\{\frac{S_T}{S_{T/2}} - 1, 0\right\}\right] \\[3pt] & = \tilde{E}\left[\max\left\{\frac{S_T}{S_{T/2}} - 1, 0\right\}\right] \end{align}$$

where we use $S_{T/2}$ as the numeraire. Then plug into the call option Black-Scholes formula with $S=1$, $K=1$ and $r=0$.

Is this approach correct? My main concern is, can we use $S_{T/2}$ as the numeraire?

  • 1
    $\begingroup$ Under the geometric Brownian motion assumption, $S_{T/2}$ and $S_T/S_{T/2}$ are independent. $\endgroup$
    – Gordon
    Oct 4, 2018 at 13:26
  • $\begingroup$ Thanks! I understand the independent property. Can you deliberate how to use this property? $\endgroup$ Oct 4, 2018 at 15:07
  • 1
    $\begingroup$ Given the independence, then $e^{-rT} E\big(\max(S_T-S_{T/2}, 0)\big)= e^{-rT} E\big(S_{T/2}\max(S_T/S_{T/2}-1, 0)\big)=e^{-rT} E\big(S_{T/2}\big)E\big(\max(S_T/S_{T/2}-1, 0)\big)$. $\endgroup$
    – Gordon
    Oct 4, 2018 at 16:44

1 Answer 1


Let us assume we are in a standard Black-Scholes setting with 2 traded assets, a money market account $B_t$ and a stock $S_t$, such that: $$\begin{align} \text{d}B_t & = rB_t\text{d}t \\ \text{d}S_t &= rS_t\text{d}t+\sigma S_t\text{d}W^Q_t \end{align}$$ where $W^Q_t$ is a Brownian motion under the risk-neutral measure associated to the bank account $B_t$. Defining $\tilde{B}_t = B_t/S_t$, by Itô's Lemma and Girsanov theorem: $$\begin{align} \text{d}\tilde{B}_t&=\sigma^2\tilde{B}_t\text{d}t-\sigma\tilde{B}_t\text{d}W^Q_t \\ &=\sigma\tilde{B}_t\left(\sigma\text{d}t-\text{d}W^Q_t\right) \\ &=\sigma\tilde{B}_t\text{d}W^S_t \end{align}$$ where $W^S_t=\sigma t-W^Q_t$ is a Brownian Motion under the measure associated to the stock numéraire, thus: $$\begin{align} \text{d}S_t & = rS_t\text{d}t+\sigma S_t\text{d}W^Q_t \\ &=(r+\sigma^2)S_t\text{d}t-\sigma S_t\text{d}W^S_t \end{align}$$ Under measure $Q^S$ the stock price is lognormal with drift $r+\sigma^2$ and volatility $\sigma$. It comes by a change of measure: $$\begin{align} E^Q_t\left[\frac{B_t}{B_T}\left(S_T-S_{T/2}\right)^+\right]&=E^Q_t\left[\frac{B_tS_T}{B_T}\left(1-\frac{S_{T/2}}{S_T}\right)^+\right] \\ &=S_tE^S_t\left[\left(1-\frac{S_{T/2}}{S_T}\right)^+\right] \end{align}$$ Yet: $$ \frac{S_{T/2}}{S_T}=\exp\left\{\sigma\sqrt{\frac{T}{2}}Z-\left(r+\frac{\sigma^2}{2}\right)\frac{T}{2}\right\}$$ where $Z \sim (-Z) \sim \mathcal{N}(0,1)$ is a standard Normal random variable. You can now recycle the Black-Scholes formula to find the price of the option.

  • $\begingroup$ Use the independence as I mentioned above, you do not need the measure change. But this can be treated as an alternative solution. $\endgroup$
    – Gordon
    Oct 4, 2018 at 13:47
  • $\begingroup$ You're right, I didn't gave it much thought, I carried on the "stock-as-numéraire" path. $\endgroup$ Oct 4, 2018 at 14:14

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